$$c_{k+1} = c_k - tan(c_k)$$ What kind of way can be used to analyze this sequence? I want to get something by calculating the difference of $c_{k+1} - c_k$ and $c_{k} - c_{k-1}$, but it turns out to be nothing that I can relate to.
what is the limit of this sequence with tangent?
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sequences-and-series
limits
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0If $C=\lim_{k\to\infty}c_k$, it follows that$$C=C-\tan(C)$$if the limits exists, of course. That's the hardest step, and it usually follows through inequalities and such, but it's not as obvious here. – 2017-02-06
2 Answers
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Let $f(x) = x - \tan(x)$. This has a stable fixed point at $x=0$ since $f(0) = 0$ and $|f'(0)| < 1$ (in fact $f'(0)=0$). Thus there is some $\epsilon > 0$ such that the limit of your sequence is $0$ if it starts out with $|c_0| < \epsilon$.
EDIT: The immediate basin of attraction of this fixed point, i.e. the largest interval containing $0$ consisting of points attracted to $0$ under this map, is $(-a,a)$ where $f(a) = -a$, $a \approx 1.165561185$. Thus $a$ and $-a$ form a $2$-cycle.
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If you use Newton's method to find the zero of $f(x)=\sin x$ you'll see that the approximations $c_k$ of the zeros are $$ c_{k+1}=c_k-\dfrac{f(c_k)}{f'(c_k)}=c_k-\tan c_k. $$ If $c_0$ is in a small neighborhood of $n\pi$ for some integer $n$, you can show that $\lim_kc_k=n\pi$.